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Mixed Numbers: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A form entry

A mixed number has whole-number part $4$ . Its fraction part is positive, has denominator $9$ , and has the largest whole-number numerator that a proper fraction allows. Write the mixed number.
1. Hint 1
  A proper fraction’s numerator is smaller than its denominator.
2. Hint 2
  List the numerators that keep a fraction with denominator $9$ proper, and write the largest one over $9$ beside the $4$ .
Answer

$4\tfrac89$ .

Full solution

The largest whole-number numerator smaller than $9$ is $8$ .
The fraction part is therefore $\frac89$ .
Together with the whole part, the number is
$4+\frac89=4\tfrac89.$
The space in the mixed number stands for addition.

Answer

$4\tfrac89$ .

Key idea

A mixed number combines its whole part with a fraction part that is smaller than one.
Problem 1: Mark complete
Problem 2 A tiled border

A border is made of $61$ tiles laid end to end in one row, and each tile is $\frac18$ of a meter long. Write the length of the border as a mixed number.
1. Hint 1
  The total length is a count of eighths of a meter, and eight eighths make one meter.
2. Hint 2
  Divide $61$ by $8$ : the quotient counts whole meters, and the remainder counts the eighths of a meter left over.
Answer

$7\tfrac58$ meters.

Full solution

The border is $61$ eighths of a meter long, which is $\frac{61}8$ meters.
Eight goes into $61$ seven times, using $56$ , and leaves $5$ , so
$\frac{61}8=7\tfrac58.$
The remainder $5$ is less than $8$ , so the fraction part is proper.
Converting back gives $7\times8+5=61$ , matching the number of tiles.

Answer

$7\tfrac58$ meters.

Key idea

The quotient of the numerator divided by the denominator counts the complete wholes, and the remainder over the same denominator becomes the fraction part.
Problem 2: Mark complete
Problem 3 A specified label

Write $3\tfrac56$ as one fraction with denominator $18$ and a whole-number numerator.
1. Hint 1
  The mixed number means three wholes plus five sixths.
2. Hint 2
  Count the sixths in the three wholes, then change the resulting fraction to eighteenths.
Answer

$\frac{69}{18}$ .

Full solution

Three wholes contain eighteen sixths, and five more sixths give
$3\times6+5=23.$
Therefore
$3\tfrac56=\frac{23}6.$
Multiply the numerator and denominator by $3$ to obtain the requested form:
$\frac{23}6=\frac{69}{18}.$
Dividing both by $3$ checks that the value was preserved.

Answer

$\frac{69}{18}$ .

Key idea

A mixed number can be counted as one improper fraction before it is rewritten with a required denominator.
Problem 3: Mark complete
Problem 4 Two walkers

Ana walks $1\tfrac58$ kilometers in the morning and $2\tfrac7{10}$ kilometers in the afternoon. Ben walks $4\tfrac18$ kilometers in one walk. Who walks farther, and by how much? Give the difference as a fraction in lowest terms.
1. Hint 1
  Ana’s two walks must be combined into one distance before it can be compared with Ben’s.
2. Hint 2
  Add Ana’s whole parts and fraction parts separately, using fortieths, and check whether the fraction parts pass one whole.
3. Hint 3
  Write Ben’s fraction part in fortieths too, then compare the two distances and subtract the smaller from the larger.
Answer

Ana, by $\frac15$ of a kilometer.

Full solution

Add Ana’s walks by parts.
The whole parts give $1+2=3$ .
In fortieths, $\frac58=\frac{25}{40}$ and $\frac7{10}=\frac{28}{40}$ , so the fraction parts give
$\frac{25}{40}+\frac{28}{40}=\frac{53}{40}.$
Since $53=1\times40+13$ , that is $1\tfrac{13}{40}$ , so carry the $1$ into the whole part: Ana walks $4\tfrac{13}{40}$ kilometers.

In fortieths, $\frac18=\frac5{40}$ , so Ben walks $4\tfrac5{40}$ kilometers.
Both distances have whole part $4$ , and $\frac{13}{40}$ is more than $\frac5{40}$ , so Ana walks farther.

The whole parts are equal, so the difference is the difference of the fraction parts:
$\frac{13}{40}-\frac5{40}=\frac8{40}.$
Dividing the top and bottom by $8$ gives $\frac15$ of a kilometer, in lowest terms.

Check with improper fractions.
Since $1\times8+5=13$ and $2\times10+7=27$ , Ana walks $\frac{13}8+\frac{27}{10}$ kilometers, which in fortieths is
$\frac{65}{40}+\frac{108}{40}=\frac{173}{40}.$
Since $4\times8+1=33$ , Ben walks $\frac{33}8$ kilometers, which is $\frac{165}{40}$ , so the difference is again $\frac8{40}$ .

Answer

Ana, by $\frac15$ of a kilometer.

Key idea

When fraction parts add past one whole, carry that whole into the whole-number part before comparing, since the whole parts alone can mislead.
Problem 4: Mark complete
Problem 5 A package adjustment

A package contains just two things: $2\tfrac38$ kilograms of packing material and $3\tfrac23$ kilograms of equipment. A worker removes $\frac78$ of a kilogram of packing material. What is the total mass of its contents now? Give a mixed number with its fraction part in lowest terms.
1. Hint 1
  Find the packing material left, then combine it with the unchanged equipment mass.
2. Hint 2
  The packing subtraction is in eighths, and the final addition can be done in sixths.
Answer

$5\tfrac16$ kilograms.

Full solution

The fraction part $\frac38$ is less than $\frac78$ , so borrow one whole: since one whole is $\frac88$ , $2\tfrac38$ becomes $1\tfrac{11}8$ .
Subtracting the fraction parts leaves
$1\tfrac{11}8-\frac78=1\tfrac48.$
This simplifies to $1\tfrac12$ kilograms of packing material.

Now add $1\tfrac12$ and $3\tfrac23$ by parts.
The whole parts give $1+3=4$ .
In sixths, the fraction parts give
$\frac36+\frac46=\frac76.$
Since $\frac76=1\tfrac16$ , carry the $1$ : the contents now total $5\tfrac16$ kilograms.

As a check, in twenty-fourths $\frac38=\frac9{24}$ and $\frac23=\frac{16}{24}$ , so the original contents are $5$ kilograms plus $\frac{25}{24}$ of a kilogram.
Removing $\frac78=\frac{21}{24}$ leaves $5$ kilograms plus $\frac4{24}$ , which is $5\tfrac16$ kilograms.

Answer

$5\tfrac16$ kilograms.

Key idea

Updating one part of a total and then combining the parts keeps the bookkeeping clear.
Problem 5: Mark complete
Problem 6 A trimmed poster

A poster is $2\tfrac25$ meters long. It is enlarged so that its new length is $1\tfrac23$ times its old length, and then $\frac12$ of a meter is trimmed from the enlarged poster. Find its final length as a mixed number in lowest terms.
1. Hint 1
  The enlargement happens before the trimming.
2. Hint 2
  Write both mixed numbers as improper fractions for the multiplication, then subtract the trimmed length.
Answer

$3\tfrac12$ meters.

Full solution

The original length is $\frac{12}5$ meters, since $2\times5+2=12$ , and the multiplier $1\tfrac23$ is $\frac53$ , since
$1\times3+2=5.$
The $5$ s cancel, and $12$ divided by $3$ is $4$ , so
$\frac{12}5\times\frac53=4.$
The enlarged poster is $4$ meters long.

Trim half a meter from that length:
$4-\frac12=\frac72.$
The final length is $3\tfrac12$ meters.
Adding the trimmed half meter returns the enlarged length of $4$ meters.

Answer

$3\tfrac12$ meters.

Key idea

Convert mixed numbers before multiplying, and apply later changes to the resulting amount.
Problem 6: Mark complete
Problem 7 A filling timer

A container already holds $\frac16$ of a liter. A pump steadily adds $2\tfrac{11}{12}$ liters each minute. How many minutes will it take to reach $4\tfrac34$ liters? Give a mixed number in lowest terms.
1. Hint 1
  The pump needs to supply the target amount minus the amount already present.
2. Hint 2
  Convert the target and the amount per minute to improper fractions, then divide the amount still needed by the amount supplied each minute.
Answer

$1\tfrac47$ minutes.

Full solution

The target is $\frac{19}4$ liters, since
$4\times4+3=19.$
In twelfths, $\frac{19}4=\frac{57}{12}$ and
$\frac16=\frac2{12}.$
The amount still needed, in liters, is
$\frac{19}4-\frac16=\frac{55}{12}.$
The pump adds $\frac{35}{12}$ liters each minute, since
$2\times12+11=35.$

The time is the needed amount divided by the amount per minute:
$\frac{55}{12}\div\frac{35}{12}=\frac{55}{12}\times\frac{12}{35}.$
The $12$ s cancel, and $\frac{55}{35}$ simplifies by $5$ , so
$\frac{55}{12}\times\frac{12}{35}=\frac{11}7.$
Since $11=1\times7+4$ , the time is $1\tfrac47$ minutes.

In that time the pump adds $\frac{11}7\times\frac{35}{12}=\frac{55}{12}$ liters.
Adding the initial $\frac16=\frac2{12}$ of a liter gives $\frac{57}{12}=4\tfrac34$ liters, the target.

Answer

$1\tfrac47$ minutes.

Key idea

The time to reach a target is found from the amount still needed, divided by the amount supplied per unit of time.
Problem 7: Mark complete
Problem 8 A card with two forms

A card shows $\frac{26}9=1+\frac{17}9$ . Explain why the right side has the correct value, then write that value as a mixed number with a proper fraction part.
1. Hint 1
  Check the value of the sum separately from the required form of its fraction part.
2. Hint 2
  Nine of the seventeen ninths form an additional whole.
Answer

$2\tfrac89$ .

Full solution

The right side has the correct value, because one whole is $\frac99$ and
$\frac99+\frac{17}9=\frac{26}9.$
But $\frac{17}9$ is improper, so the right side does not yet have a proper fraction part.

The seventeen ninths contain one whole and eight more ninths:
$\frac{17}9=1+\frac89.$
Combining that extra whole with the existing whole gives $2\tfrac89$ .
The fraction part is now proper, so the rewritten form is $2\tfrac89$ .

Answer

$2\tfrac89$ .

Key idea

A correct whole-plus-fraction expression may still need a whole moved out of its fraction part.
Problem 8: Mark complete
Problem 9 Rae’s comparison

Rae says $7\tfrac46$ and $7\tfrac23$ cannot be equal, because when each is written as an improper fraction the two results have different denominators. Is she correct? Convert both and explain.
1. Hint 1
  Rae’s reason is about how the two fractions are written. Decide what actually shows whether two fractions name the same amount.
2. Hint 2
  Convert each with the whole part times the denominator plus the numerator, then check whether each result is in lowest terms.
Answer

No; $7\tfrac46=\frac{46}6$ , which is $\frac{23}3$ , and $7\tfrac23=\frac{23}3$ .

Full solution

Converting keeps each mixed number’s denominator.
For the first, $7\times6+4=46$ , so
$7\tfrac46=\frac{46}6.$
For the second, $7\times3+2=23$ , so
$7\tfrac23=\frac{23}3.$
Rae is right that these two results have different denominators.

But $46$ and $6$ share the factor $2$ , and dividing both by $2$ gives
$\frac{46}6=\frac{23}3.$
The two mixed numbers name the same value, so Rae is not correct.
Different denominators in unsimplified forms do not mean different values.

Answer

No; $7\tfrac46=\frac{46}6$ , which is $\frac{23}3$ , and $7\tfrac23=\frac{23}3$ .

Key idea

Converting keeps the denominator, but simplifying afterward can change it, so one value can appear with different denominators.
Problem 9: Mark complete
Problem 10 Theo’s reading

Theo reads $9\tfrac45$ as $9\times\frac45$ and says its value is $7\tfrac15$ . Is he correct? Explain, and give the value of $9\tfrac45$ as an improper fraction.
1. Hint 1
  Ask what the space between the whole part and the fraction part of a mixed number stands for.
2. Hint 2
  Compare where $9\tfrac45$ must sit among the whole numbers with the size of $9$ times a number less than $1$ .
3. Hint 3
  To convert, count the fifths in nine wholes, then add the four more fifths.
Answer

No; $9\tfrac45=\frac{49}5$ .

Full solution

Theo’s reading gives
$9\times\frac45=\frac{36}5.$
Since $36=7\times5+1$ , that is $7\tfrac15$ , so his arithmetic is right but his reading is not.

The space in a mixed number means add, so $9\tfrac45$ stands for $9+\frac45$ , a value between $9$ and $10$ .
Nine wholes are $9\times5=45$ fifths, and four more fifths make $49$ , so
$9\tfrac45=\frac{49}5.$

A proper fraction is less than $1$ , so $9$ times it is less than $9$ .
The mixed number is more than $9$ , so no such product can equal it, and Theo is not correct.

Answer

No; $9\tfrac45=\frac{49}5$ .

Key idea

The space in a mixed number means add, so its value lies between its whole part and the next whole number.
Problem 10: Mark complete

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